Whakaoti i te 80=piD^2/4

$$80 = \frac { \pi D ^ { 2 } } { 4 }$$

Whakaoti mō D

$$D=8\sqrt{\frac{5}{\pi }}\approx 10.092530088$$

$$D=-8\sqrt{\frac{5}{\pi }}\approx -10.092530088$$

Ngā Upane mā te Kimi Pūtakerua

Me whakarea ngā taha e rua ki te 4.

$$80\times 4=\pi D^{2}$$

Whakareatia te 80 ki te 4, ka 320.

$$320=\pi D^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\pi D^{2}=320$$

Whakawehea ngā taha e rua ki te \pi .

$$\frac{\pi D^{2}}{\pi }=\frac{320}{\pi }$$

Mā te whakawehe ki te \pi ka wetekia te whakareanga ki te \pi .

$$D^{2}=\frac{320}{\pi }$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$D=\frac{40}{\sqrt{5\pi }} D=-\frac{40}{\sqrt{5\pi }}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua ki te 4.

$$80\times 4=\pi D^{2}$$

Whakareatia te 80 ki te 4, ka 320.

$$320=\pi D^{2}$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\pi D^{2}=320$$

Tangohia te 320 mai i ngā taha e rua.

$$\pi D^{2}-320=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi \pi mō a, 0 mō b, me -320 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$D=\frac{0±\sqrt{0^{2}-4\pi \left(-320\right)}}{2\pi }$$

Pūrua 0.

$$D=\frac{0±\sqrt{-4\pi \left(-320\right)}}{2\pi }$$

Whakareatia -4 ki te \pi .

$$D=\frac{0±\sqrt{\left(-4\pi \right)\left(-320\right)}}{2\pi }$$

Whakareatia -4\pi ki te -320.

$$D=\frac{0±\sqrt{1280\pi }}{2\pi }$$

Tuhia te pūtakerua o te 1280\pi .

$$D=\frac{0±16\sqrt{5\pi }}{2\pi }$$

Nā, me whakaoti te whārite D=\frac{0±16\sqrt{5\pi }}{2\pi } ina he tāpiri te ±.

$$D=\frac{40}{\sqrt{5\pi }}$$

Nā, me whakaoti te whārite D=\frac{0±16\sqrt{5\pi }}{2\pi } ina he tango te ±.

$$D=-\frac{40}{\sqrt{5\pi }}$$

Kua oti te whārite te whakatau.

$$D=\frac{40}{\sqrt{5\pi }} D=-\frac{40}{\sqrt{5\pi }}$$